Chemistry: Post your doubts here!

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Can someone explain why the answer is C ?
If you draw the propene molecule, it will help. so the molecule looks something like this: CH2=CH-CH3. Now we have Br2 which is (i) Br-Br and HBr which is (ii) H-Br. In case (i) the nucleophile will attack. the double bond will break and one extra bond will be available on C1 and one more on C2. a Br ion can attach to each one of these free bonds, thus forming option D. in (ii) H can bond to first free bond or the second. Br will attach to the one left. Through this, the process forms option A and B.

Option C is impossible because it implies that both the bromine ions went to the first carbon, but that is not possible because if you revisit molecular bonding, a pi bond is basically an unhybridised p orbital from each atom interacting. SO the breaking of a double bond will automatically free up one unhybridised p orbital on either C atom. You just can't have option C. Others are possible, as described above.

Hope this helps:)
 
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If you draw the propene molecule, it will help. so the molecule looks something like this: CH2=CH-CH3. Now we have Br2 which is (i) Br-Br and HBr which is (ii) H-Br. In case (i) the nucleophile will attack. the double bond will break and one extra bond will be available on C1 and one more on C2. a Br ion can attach to each one of these free bonds, thus forming option D. in (ii) H can bond to first free bond or the second. Br will attach to the one left. Through this, the process forms option A and B.

Option C is impossible because it implies that both the bromine ions went to the first carbon, but that is not possible because if you revisit molecular bonding, a pi bond is basically an unhybridised p orbital from each atom interacting. SO the breaking of a double bond will automatically free up one unhybridised p orbital on either C atom. You just can't have option C. Others are possible, as described above.

Hope this helps:)
Thank you so much!!!!
 
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